Why is 2 * (i * i) faster than 2 * i * i in Java?

Question

The following Java program takes on average between 0.50 secs and 0.55 secs to run:

public static void main(String[] args) {
    long startTime = System.nano         


        

Answer 1

I tried a JMH using the default archetype: I also added an optimized version based on Runemoro's explanation.

@State(Scope.Benchmark)
@Warmup(iterations = 2)
@Fork(1)
@Measurement(iterations = 10)
@OutputTimeUnit(TimeUnit.NANOSECONDS)
//@BenchmarkMode({ Mode.All })
@BenchmarkMode(Mode.AverageTime)
public class MyBenchmark {
  @Param({ "100", "1000", "1000000000" })
  private int size;

  @Benchmark
  public int two_square_i() {
    int n = 0;
    for (int i = 0; i < size; i++) {
      n += 2 * (i * i);
    }
    return n;
  }

  @Benchmark
  public int square_i_two() {
    int n = 0;
    for (int i = 0; i < size; i++) {
      n += i * i;
    }
    return 2*n;
  }

  @Benchmark
  public int two_i_() {
    int n = 0;
    for (int i = 0; i < size; i++) {
      n += 2 * i * i;
    }
    return n;
  }
}

The result are here:

Benchmark                           (size)  Mode  Samples          Score   Score error  Units
o.s.MyBenchmark.square_i_two           100  avgt       10         58,062         1,410  ns/op
o.s.MyBenchmark.square_i_two          1000  avgt       10        547,393        12,851  ns/op
o.s.MyBenchmark.square_i_two    1000000000  avgt       10  540343681,267  16795210,324  ns/op
o.s.MyBenchmark.two_i_                 100  avgt       10         87,491         2,004  ns/op
o.s.MyBenchmark.two_i_                1000  avgt       10       1015,388        30,313  ns/op
o.s.MyBenchmark.two_i_          1000000000  avgt       10  967100076,600  24929570,556  ns/op
o.s.MyBenchmark.two_square_i           100  avgt       10         70,715         2,107  ns/op
o.s.MyBenchmark.two_square_i          1000  avgt       10        686,977        24,613  ns/op
o.s.MyBenchmark.two_square_i    1000000000  avgt       10  652736811,450  27015580,488  ns/op

On my PC (Core i7 860 - it is doing nothing much apart from reading on my smartphone):

• n += i*i then n*2 is first
• 2 * (i * i) is second.

The JVM is clearly not optimizing the same way than a human does (based on Runemoro's answer).

Now then, reading bytecode: javap -c -v ./target/classes/org/sample/MyBenchmark.class

• Differences between 2*(i*i) (left) and 2*i*i (right) here: https://www.diffchecker.com/cvSFppWI
• Differences between 2*(i*i) and the optimized version here: https://www.diffchecker.com/I1XFu5dP

I am not expert on bytecode, but we iload_2 before we imul: that's probably where you get the difference: I can suppose that the JVM optimize reading i twice (i is already here, and there is no need to load it again) whilst in the 2*i*i it can't.